随机生成4个代表扑克牌牌面的数字字母，程序自动列出所有可能算出24的表达式，用擅长的语言（C/C++/Java或其他均可）实现程序解决问题。
1.程序风格良好(使用自定义注释模板)
2.列出表达式无重复。
以下为源代码
#include<iostream>
#include<string>
#include <stdlib.h>
#include<time.h>
using namespace std;
char card[] = { 'A','2','3','4','5','6','7','8','9','10','J','Q','K' };
char buf[4];
double nums[4];
char ope[4] = { '+','-','*','/' };
void cre()//生成
{
int i = 0;
int j;
cout << "生成的四张牌面为：";
srand((unsigned)time(0));
for (i = 0; i<4; i++)
{
j =rand() % 13 + 1;
buf[i] = card[j];
}
cout << buf[0] << ";" << buf[1] << ";" << buf[2] << ";" << buf[3] << "。" << endl;
for (i = 0; i<4; i++)
{
if (buf[i] == 'A') nums[i] = 1;
else if(buf[i] == '2') nums[i] = 2;
else if (buf[i] == '3') nums[i] = 3;
else if (buf[i] == '4') nums[i] = 4;
else if (buf[i] == '5') nums[i] = 5;
else if (buf[i] == '6') nums[i] = 6;
else if (buf[i] == '7') nums[i] = 7;
else if (buf[i] == '8') nums[i] = 8;
else if (buf[i] == '9') nums[i] = 9;
else if (buf[i] == '10') nums[i] = 10;
else if (buf[i] == 'J') nums[i] = 11;
else if (buf[i] == 'Q') nums[i] = 12;
else if (buf[i] == 'K') nums[i] = 13;
}
}
double calcute(double a, double b, char index)
{
if (index == '+') return a + b; //若为+，则返回相应结果
else if (index == '-') return a - b;
else if (index == '*') return a*b;
else if (index == '/')
if (b != 0)
return a / b; //只有当分母不为0时，返回结果
}
void exh()//穷举计算
{
double temp[3], tem[2]; //第一个符号放置后，经过计算后相当于剩下三个数，这个数组用于存储这三个数
double sum; //求得的和
int judge = 0; //判断是否找到一个合理的解
for (int i = 0; i < 4; i++) //第一次放置的符号
{
for (int j = 0; j < 4; j++) //第二次放置的符号
{
for (int k = 0; k < 4; k++) //第三次放置的符号
{
for (int m = 0; m < 3; m++) //首先计算的两个相邻数字，共有3种情况，相当于括号的作用
{
if (nums[m + 1] == 0 && ope[i] == '/') break;
temp[m] = calcute(nums[m], nums[m + 1], ope[i]);
temp[(m + 1) % 3] = nums[(m + 2) % 4];
temp[(m + 2) % 3] = nums[(m + 3) % 4]; //先确定首先计算的两个数字，计算完成相当于剩下三个数，按顺序储存在temp数组中
for (int n = 0; n < 2; n++) //三个数字选出先计算的两个相邻数字，两种情况，相当于第二个括号
{
if (temp[n + 1] == 0 && ope[j] == '/') break;
tem[n] = calcute(temp[n], temp[n + 1], ope[j]);
tem[(n + 1) % 2] = temp[(n + 2) % 3]; //先确定首先计算的两个数字，计算完成相当于剩下两个数，按顺序储存在temp数组中
if (tem[1] == 0 && ope[k] == '/') break;
sum = calcute(tem[0], tem[1], ope[k]); //计算和
if (sum == 24) //若和为24
{
judge = 1; //判断符为1，表示已求得解
if (m == 0 && n == 0)
cout << "((" << nums[0] << ope[i] << nums[1] << ")" << ope[j] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
else if (m == 0 && n == 1)
cout << "(" << nums[0] << ope[i] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[j] << nums[3] << ")=" << sum << endl;
else if (m == 1 && n == 0)
cout << "(" << nums[0] << ope[j] << "(" << nums[1] << ope[i] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
else if (m == 1 && n == 1)
cout << nums[0] << ope[k] << "((" << nums[1] << ope[i] << nums[2] << ")" << ope[j] << nums[3] << ")=" << sum << endl;
else if (m == 2 && n == 0)
cout << "(" << nums[0] << ope[j] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[i] << nums[3] << ")=" << sum << endl;
else if (m == 2 && n == 0)
cout << nums[0] << ope[k] << "(" << nums[1] << ope[j] << "(" << nums[2] << ope[i] << nums[3] << "))=" << sum << endl; //m=0,1,2 n=0,1表示六种括号放置可能，并按照这六种可能输出相应的格式的计算式
}
}
}
}
}
}
if (judge == 0)
cout << "这四张扑克牌无法找到一个合理的解" << endl; //如果没有找到结果，符号位为0
}
int main()
{
int i;
int select = 1;
cout<< " ################################################" << endl
<< " # #" << endl
<< " # 欢迎进入24点游戏 #" << endl
<< " # #" << endl
<< " ################################################" << endl;
while (select)
{
cout<< " ################################################" << endl
<< " # #" << endl
<< " # 是否开始游戏 #" << endl
<< " # #" << endl
<< " # 0.是 1.否 #" << endl
<< " # #" << endl
<< " ################################################" << endl;
cout << "请输入你的选择（0或1）：";
cin >> i;
switch (i)
{
case 0:
cre();
exh();
break;
case 1:
select = 0;
break;
default:
cout << "请在0和1之间选择！" << endl;
}
}
return 0;
}
